Attempt: Let $X$ denote the number of boxes without balls. This means $$ EX = \sum_{x=0}^{4} x P(X=x) $$
Since we know each ball must go into at least one box, we cannot have 5 empty boxes, so that is why I sum to 4.
I then said $P(X=j) = P(j{}\,\text{boxes with no balls})= {5 \choose j}(1-p_i)^{10}$ So $$EX = 0 + \sum_{j=1}^{4} j{5 \choose j}(1-p_i)^{10}$$ Is it ok?

You could maybe add that you are using Bernoulli RV, may not be clear from is point of view
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Jean-SébastienDec 6 '12 at 16:04

Is there a way to get the right answer using my method? Instead of finding the prob that a particular box was empty, I found the prob that $i$ boxes was empty.
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CAFDec 6 '12 at 16:21

@CAF: You can certainly do what you describe in this comment, but that's pretty far from what you write in the question. You can't use $\binom5j$ because you'd have to multiply that with some probability that's the same for all selections of $j$ empty boxes, but the boxes all have different probabilities of being empty. So if you want to do it as you describe in the comment, you have to add up all the individual contributions; doing that and then bring that into the simple form $\sum_i(1-p_i)^{10}$ is going to be quite a pain.
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jorikiDec 6 '12 at 16:42

Ok, thanks. The next question was find the expected number of boxes with 1 ball in it. So the prob that one box has one ball in it is $ {10 \choose 1} (p_i) (1-p_i)^9 $ so then I just sum this from 1 to 5?
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CAFDec 6 '12 at 16:51

@CAF: Exactly. Note that if you also do this for $2,\dotsc,10$ balls per box and take the sum of all these expected numbers, you can then swap the two sums and evaluate the inner sum to $1$, and then the outer sum yields $5$ as it must.
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jorikiDec 6 '12 at 17:28